Question

Give an example of extension field K and L of F such that both K and L are Galois over $\mathit{F}\mathbf{,}\mathit{K}\mathbf{\ne }\mathit{L}$, and role="math" localid="1657963027377" $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathbf{\cong }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{L}$.

Short answer

$\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathbf{\cong }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{L}$

Step-by-step solution

Definition of Galois group.

The set of all F automorphisms ofK is an extension ofF then the Galois group is denotes by $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$ it is the Galois group of $\mathit{K}$ over $\mathit{F}$.

Step-2: Showing that GalF≅GalFL .

$$\begin{gathered} \mathit{\sigma }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathbf{a}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{b} \end{gathered}$$

Consider the extension field K and L over F are Galois field over F but both field are not same since the element of the both field are different and Galois group of field are same $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}\mathbf{\cong }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{L}$.

Let two element of the extension field $\mathit{K}$ is $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{K}$and another two element of the extension field are $\mathit{a}\mathbf{,}\mathit{B}\mathbf{\in }\mathit{L}$ and role="math" localid="1657963284874" $\mathit{K}$ is automorphism over the both extension field is depend on the isomorphism since the isomorphism of the field is

For role="math" localid="1657963458362" $\mathit{K}$

$\mathit{\varphi }\left(a\right)\mathbf{=}\mathit{b}$

For L

$\mathit{F}\mathit{\varphi }\left(a\right)\mathbf{=}\mathit{B}$

Further the $\mathit{K}$ is automorphism over the extension field $\mathit{K}$.

$$\begin{gathered} \mathit{\sigma }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathbf{a}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{b} \end{gathered}$$

Further the $\mathit{K}$automorphism over the extension field $\mathit{L}$.

$$\begin{gathered} \mathit{\sigma }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathbf{a}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{B} \end{gathered}$$

Since the Galois group of the both extension field are same $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathbf{\cong }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{L}$ but the field are not same consist the element but degee and order are same. Hence extension field are not equal $\mathit{K}\mathbf{\ne }\mathit{L}$.